# Suppose $L$ has a regular parametrix . Assume $U$ is a distribution given in an open set $\Omega \subset R^d$ and $L(U)=f$ , then $U$ is $C^{\infty}$

Suppose $$L$$ has a regular parametrix . Assume $$U$$ is a distribution given in an open set $$\Omega \subset R^d$$ and $$L(U)=f$$ , with $$f$$ a $$C^{\infty}$$ function in $$\Omega$$ , then $$U$$ agrees with a $$C^{\infty}$$ function on $$\Omega$$

I have learned this theorem in Stein's functional analysis Page$$_{133}$$ and the author proved this theorem in the following step .

$$(1)$$ Show that it suffices to prove $$U$$ agrees with a $$C^\infty$$ function on any ball $$B$$ with its closure $$B^* \subset \Omega$$ .

$$(2)$$ Show that for any ball $$B$$ with its closure $$B^* \subset \Omega$$ , we can find a function $$f_B \in C^{\infty}$$ which depends on the ball $$B$$ such that $$U$$ agree with $$f_B$$ on $$B$$ .

After finished the above two step , the author state that the theorem is proved , but I did not see it .

My attempt :
For step $$(1)$$ , assume $$U$$ argees with $$g$$ . We need to prove for all $$\varphi \in D(\Omega)$$ , we have $$U(\varphi)=\int g(x)\varphi(x) \,dx$$ If this is not true for some $$\varphi$$ . We can find a function $$\phi$$ such that $$L\phi=\varphi$$ . Then $$\int g(x)\varphi(x) \,dx \neq U(\varphi)=L(U)(\phi)=\int f(x) \phi(x) \,dx$$ WLOG , we have $$g(x)\varphi(x) \lt f(x) \phi(x)$$ in some open ball $$O$$ . Then $$\int g(x) \varphi(x) \chi_O(x) \,dx \lt \int f(x)\phi(x) \chi_O(x) \,dx$$
Notice that $$\varphi(x)\chi_O(x)=L(U)(\phi(x)\chi_O(x))$$ . However , these two functions are not in $$C^\infty$$ . So I want to show $$U(\varphi(x)\chi_O(x))=\int g(x) \varphi(x)\chi_O(x) \,dx$$ and $$L(U)(\varphi(x)\chi_O(x))=\int f(x) \varphi(x)\chi_O(x) \,dx$$

My question :
$$(a)$$ Can we show step $$(1)$$ follow the discussion above ? If we do not have the assumption $$L(U)=f$$ , does $$(1)$$ still valid ?

$$(b)$$ In part $$(2)$$ , for each ball $$B$$ , we have different functions $$f_B \in C^{\infty}(\Omega)$$ , so how to use $$(2)$$ to show $$U$$ agrees with a function $$f$$ ?

For each ball $$B_{\alpha}$$ , let $$f_{\alpha}$$ denote the function which agrees with distribution $$U$$ on $$B_{\alpha}$$ . Since $$\Omega$$ is open , it can be write as a union (might not disjoint) of open balls $$\bigcup B_{\alpha}$$ . We now define the function $$f$$ as follows .
For each $$x$$ , we can find a ball $$B_{\alpha}$$ contains $$x$$ ,and we define $$f(x)=f_{\alpha}(x)$$ . To see $$f$$ is well defined , WLOG, assume $$B_{\beta}$$ is another ball contains $$x$$ and $$f_{\beta}(x)\lt f_{\alpha}(x)$$ . Then we can find a ball $$C$$ contains $$x$$ with $$C \subset B_{\alpha}$$ and $$C\subset B_{\beta}$$ such that whenever $$t \in C$$ we have $$f_{\beta}(t)\lt f_{\alpha}(t)$$ . Find a positive function $$\phi \in D(\Omega)$$ with $$\phi(x)\gt 0$$ and supported in $$C$$ , then we have $$U(\phi)=\int f_{\alpha}(t)\phi(t) \, dt=\int f_{\beta}(t)\phi(t) \, dt$$ which is impossible . So the function $$f$$ is well defined . Also , note that for each $$x\in \Omega$$ , we can find a ball $$B_{\alpha}$$ contains $$x$$ , and $$f(x)=f_{\alpha}(x)$$ whenever $$x \in B_{\alpha}$$ . So $$f \in C^{\infty}(\Omega)$$ .
To see $$F(\varphi)=\int f(x)\varphi(x)\,dx$$ whenever $$\varphi \in D(\Omega)$$ . Since the support $$K$$ of $$\varphi$$ is compact , we can find finite union of balls to cover $$K$$ , we say $$K\subset \bigcup_{n=1}^N B_n$$ . Then by the partition of unity , we can find function $$\eta_k$$ $$(1\le k\le N)$$ $$\in D$$ with $$supp(\eta_k)\subset B_k$$ and $$\sum_{n=1}^N\eta_n(x)=1$$ whenever $$x\in K$$ . Then we have $$F(\varphi)=F(\sum_{n=1}^N \varphi(x)\eta_n(x))=\sum_{n=1}^N F( \varphi(x)\eta_n(x))=\sum_{n=1}^N \int \varphi(x) \eta_n(x) f(x) \,dx \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\int f(x)\varphi(x) \,dx$$ And the proof is completed .